forced pendulum

In a review lectures Feynman gave to his freshman students, just before their first big exam, he advised them as follows (copied from Feynman's Tips on Physics, a problem-solving supplement to The Feynman Lectures on Physics):

"Now, all these things you can feel. You don’t have to feel them; you can work them out by making diagrams and calculations, but as problems get more and more difficult, and as you try to understand nature in more and more complicated situations, the more you can guess at, feel, and understand without actually calculating, the much better off you are! So that’s what you should practice doing on the various problems: when you have time somewhere, and you’re not worried about getting the answer for a quiz or something, look the problem over and see if you can understand the way it behaves, roughly, when you change some of the numbers."

Solve the problem given below (originally homework for FLP Vol. I, chapter 23) in the spirit of Feynman's advice, above. It must be solved without using any calculus or differential equations or integral equations or difference equations, etc., without iterative numerical methods, nor any such fancy mathematical tricks! You may use only algebra, geometry, trigonometry, dimensional analysis, and Newtonian mechanics, in your solution, which should be guided by your physical intuition (however note: all intuitions used in solutions must be justified)! Your answer does not have to be exact, but it should at least be a very close approximation. And be sure to show all your work! Here is the problem:

The pivot point of a simple pendulum having a natural period of 1.00 second is moved laterally in a sinusoidal motion with an amplitude 1.00 cm and period 1.10 seconds. With what amplitude should the pendulum bob swing after a steady motion is attained?

Solution by Ruggero Altair

By definition the natural period is the period of the simple pendulum when the pivot is at rest. That, by dimensional analysis (using the fact that we have a unique way to get a time out of a simple pendulum dimensional quantities), gives the following relation between the period and the length of the simple pendulum:

T = c (l/g)^(1/2) = 1.0s where g is gravity, l is the length of the simple pendulum (that we don't know) and c is a constant we also don't know (at least not without using calculus or some experimental evidence).

Now we know from the problem that the system reaches a steady motion, and we argue that the steady motion, to be steady, has to be in phase with the sinusoidal motion of the pivot point that is forced to oscillate with period T'=1.1s, which means that when the oscillating pivot point is at its maximum right the pendulum is also at its maximum right (actually it could also be in counter phase if the pivot was moving with a period T' shorter than T). (the fact that being in phase or counter phase is a requirement of the steady motion is quite obvious, but just to make it explicit it's because by definition of a steady motion we don't want a motion with secular effects or quasi periodic effects).

Assuming that we are in the small angle approximation (which we can't check right away but we have to assume if we are using in any meaningful way the definition of natural period), we can see that a motion in phase with the pivot point looks like a longer simple pendulum of which we can just see the inferior part (this is not true in general, since the length of our pendulum doesn't change while oscillating, while in the longer pendulum I'm describing the length of the part below a certain line would actually be longer when the pendulum is at it's minimum height than when it is at its maximum height, so this is why we need the small angle approximation and we need to consider the length as being constant).

This means that our oscillating pivot point is just mimicking a "half-a-way" point of this longer simple pendulum (in the small angle approximation).

Now, because we want the pendulum to be in phase with the pivot point, this implies that the natural period of this longer pendulum is the same as the pivot point period, which is T' = 1.1s. But T', as we just explained, has to be also the period of the longer pendulum, which is, again by dimensional analysis:

T' = c (l'/g)^(1/2) where l' is the length of the longer pendulum, and again, we don't know l' and c.

Now, because we are looking at the (longer) pendulum in the small angle approximation, we can look at two similar triangles defined by the (longer) pendulum itself.

One triangle is defined by the length of the longer pendulum and the distance from the vertical (with respect of the imaginary pivot point of the longer simple pendulum itself). The other triangle is defined by the distance of the horizontal oscillating pivot point and the imaginary fixed pivot point, and the amplitude of the oscillating pivot point (half of it).

Being the ratio of the two sides of these two triangles the same by euclidean geometry, we can write the following proportion:

l'/(A'/2) = (l'-l)/(A/2) where A is the amplitude of the oscillating pivot point, l'-l is the distance between the imaginary pivot point and the oscillating pivot point at it's maximum position, l' is the length of the imaginary longer pendulum, and A' is the amplitude we are supposed to calculate.

Now, from the expressions of the periods given above, we can just plug in and get an answer for A', which is

A' = A (T')^2/[(T')^2-(T)^2] = (1cm) (1.1s)^2/[(1.1s)^2-(1s)^2] = 5.76 cm

which is the answer we were looking for.

Now, three things are worth noting:

1) Even if used in the formulae, we never actually needed the values l, l', c, g, which would not change the answer as long as we stay in the small angle approximation;

2) from simple experimental observations, we do know what the constant c written above is, and that actually allows us to check that we really are in the small angle approximation for such periods and amplitudes (or lengths);

3) if the period of the forced oscillation of the pivot was shorter than the natural period of the simple pendulum, then the above procedure could still be applied to get the solution, as long as we consider the counter phase movement of the pendulum (which is, when the pivot is at its maximum right, the pendulum is at its maximum left). This, also in the small approximation, would create an imaginary pivot at a point lower than the horizontal oscillating pivot, allowing us to use different lengths for the period and find a shorter period.